Tham khảo

\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x.\left(x+3\right)}=\frac{667}{2002}\)

\(=\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{x.\left(x+3\right)}\right)=\frac{667}{2002}\)

\(=\frac{1}{3}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{667}{2002}\) 

\(=\frac{1}{3}.\left(\frac{1}{1}-\frac{1}{x+3}\right)=\frac{667}{2002}\) 

                  \(\frac{1}{1}-\frac{1}{x+3}=\frac{667}{2002}:\frac{1}{3}\)

                   \(\frac{1}{1}-\frac{1}{x+3}=\frac{2001}{2002}\) 

                              \(\frac{1}{x+3}=1-\frac{2001}{2002}\) 

                               \(\frac{1}{x+3}=\frac{1}{2002}\) 

                                \(\frac{1}{x}=\frac{1}{2002-3}\) 

                                 \(\frac{1}{x}=\frac{1}{1999}\)

Vậy x = 1999

a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x.\left(x+1\right)}=\frac{667}{668}\)

\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{667}{668}\)

\(1-\frac{1}{x+1}=\frac{667}{668}\)

\(\frac{1}{x+1}=1-\frac{667}{668}\)

\(\frac{1}{x+1}=\frac{1}{668}\)

\(\Rightarrow x+1=668\)

x = 667

a) 1/1x2 + 1/2x3 + 1/3x4 + ... + 1/x.(x+1) = 667/668

=>1/1-1/2+1/2-1/3+1/3-1/4+.......+1/x-1/x+1=667/668

=>1/1-1/x+1=667/668

=>1/x+1=1/1-667/668

=>1/x+1=1/668

=>x=667

b) \(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x.\left(x+3\right)}=\frac{667}{2002}\)

\(\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{667}{2002}\)

\(\frac{1}{3}.\left(1-\frac{1}{x+3}\right)=\frac{667}{2002}\)

\(1-\frac{1}{x+3}=\frac{2001}{2002}\)

\(\frac{1}{x+3}=\frac{1}{2002}\)

=> x + 3 = 2002

x = 1999

\(\frac{x-1}{2000}+\frac{x-3}{1998}+\frac{x-5}{1996}+\frac{x}{667}=6\)

\(\Rightarrow\frac{x-1}{2000}+\frac{x-3}{1998}+\frac{x-5}{1996}+\frac{x}{667}-6=0\)

\(\Rightarrow\left(\frac{x-1}{2000}-1\right)+\left(\frac{x-3}{1998}+1\right)+\left(\frac{x-5}{1996}-1\right)+\left(\frac{x}{667}-3\right)=0\)

\(\Rightarrow\frac{x-1-2000}{2000}+\frac{x-3-1998}{1998}+\frac{x-5-1996}{1996}+\frac{x-3.667}{667}=0\)

\(\Rightarrow\frac{x-2001}{2000}+\frac{x-2001}{1998}+\frac{x-2001}{1996}+\frac{x-2001}{667}=0\)

\(\Rightarrow\left(x-2001\right)\left(\frac{1}{2000}+\frac{1}{1998}+\frac{1}{1996}+\frac{1}{667}\right)=0\)

Ta có: \(\frac{1}{2000}+\frac{1}{1998}+\frac{1}{1996}+\frac{1}{667}\ne0\)

\(\Rightarrow x-2001=0\Rightarrow x=2001\)

667 - 44 - x = 67 

623 - x       = 67

x               = 623 - 67 

x               = 556 

\(667-\left(44-x\right)=67\)

\(=>x=-556\)

667 - [ 44 - x ] =67

        [ 44 - x ] = 667 - 67

          44 - x  = 600

                x   = 44 - 600

                x   = - 556